For gas-pressure-supported stellar homology with specific nuclear heating $ϵ={ϵ}_0\rho T^{13}$ and opacity $\kappa ={\kappa }_0\rho T^{-3}$, the dimensionless radiative coefficient is proportional to $L{\mu }^{-7}{M}^{-5}$. The nuclear coefficient is proportional to ${\mu }^{13}{M}^{15}/(L{R}^{16})$. Holding both and the dimensionless profiles fixed gives the displayed scalings. The Stefan–Boltzmann law then gives $T_e\propto {\mu }^{25/16}{M}^{15/16}$ and $L\propto {\mu }^{-4/3}{T}_e^{16/3}$. The coefficients, pressure regime and composition assumptions must be held as specified; these relations are not universal laws for massive stars.

For a chemically uniform, ideal-gas-supported stellar homology sequence with nuclear energy generation $ϵ={ϵ}_0{X}^2\rho T^4$ and opacity $\kappa ={\kappa }_{0}Z(1+X)\rho T^{-4}$, radiative diffusion in a star fixes $LRZ(1+X)/({\mu }^8{M}^6)$ and nuclear heating fixes $X^2{\mu }^4{M}^6/(R^{7}L)$. Eliminating $L$ gives $R^6\propto X^{2}Z(1+X){\mu }^{-4}$, henceThe absence of $M$ in the radius law is a consequence of these particular power laws; it is not a universal relation for real stars. At fixed initial $X$ and $\mu$, the Stefan–Boltzmann law gives $T_e\propto Z^{-1/12}$ at fixed luminosity.

For gas-pressure-supported stellar homology with opacity $\kappa ={\kappa }_{0}Z\rho T^{-3}$ and stellar energy-generation rate $ϵ={ϵ}_0{X}^2\rho T^5$, radiative diffusion in a star gives $L\propto R{T}_c^7/(Z{\rho }_c^2)\propto {\mu }^7{M}^5/Z$. Integrating the heating rate instead gives $L\propto X^2{\mu }^5{M}^7/R^8$. Equality in stellar thermal equilibrium yields the displayed radius law and $T_c\propto {\mu }^{5/4}{M}^{3/4}{X}^{-1/4}{Z}^{-1/8}$. This fixes the dimensionless profiles and coefficients, and neglects radiation pressure; it is a model comparison rather than a universal law for all main sequence stars. At fixed mass and metal mass fraction, homogeneous hydrogen burning gives the homogeneous fuel-depletion luminosity feedback through $L\propto {\mu }^7$.

For a chemically uniform, gas-pressure-supported stellar homology with Kramers' opacity law and specific heating $ϵ={ϵ}_0\rho T^{\nu }$, radiative diffusion in a star gives $L\propto {\kappa }_0^{-1}{\mu }^{15/2}{M}^{11/2}{R}^{-1/2}$. Nuclear heating gives $L\propto {ϵ}_0{\mu }^{\nu }{M}^{\nu +2}{R}^{-\nu -3}$. Equating them yields $R^{\nu +5/2}\propto {ϵ}_0{\kappa }_0{\mu }^{\nu -15/2}{M}^{\nu -7/2}$. Thus $\nu =7/2$ gives a mass-independent radius, whereas $\nu =23/2$ gives $R\propto M^{4/7}$ and $L\propto M^{73/14}$. The coefficients and dimensionless profiles must be fixed; this is an idealized radiative model rather than a universal stellar law.

For gas-pressure-supported stellar homology with $\kappa \propto {\rho }^{1/2}{T}^{-5/2}$ and specific nuclear heating $ϵ\propto \rho T^5$, the central scalings give $L_{\mathrm{nuc}}\propto M^7{R}^{-8}$ and radiative diffusion in a star gives $L_{\mathrm{rad}}\propto M^5{R}^{-1}$. Equating them yields the displayed laws. The Stefan–Boltzmann law gives $T_{\mathrm{eff}}\propto M^{29/28}$, hence $d\log L/d\log T_{\mathrm{eff}}=132/29$. Composition and dimensionless profiles are held fixed.

The antipodal map on the two-sphere reverses orientation, so its mapping torus is a nonorientable closed three-manifold. The Wang sequence gives $H_0=H_1=\mathbb{Z}$, $H_2=\mathbb{Z}/2$ and $H_3=0$. The universal coefficient theorem for cohomology therefore gives the cohomology ring $\mathbb{Z}[t,u]/(t^2,tu,u^2,2u)$ with $|t|=1$, $|u|=3$. The base-circle class satisfies $t^2=0$ by pullback; the other positive-degree products vanish for dimension reasons.

The singular chain complex of a product is naturally chain-homotopy equivalent to the tensor product of the factors' chain complexes. The shuffle map and the front-face/back-face map implement the equivalence. It is the chain-level input to the Künneth theorem.

The density-weighted mean kinematic viscosity of the universal vertical profile is $\bar{\nu }=\alpha \Omega h^2{\int }_{-{\zeta }_s}^{{\zeta }_s}Pd\zeta$. Substitution of the height scale gives $\bar{\nu }=C{\alpha }^{4/3}(\kappa /\sigma {)}^{1/3}{\mathcal{R}}^{4/3}{\Sigma }^{2/3}{\Omega }^{-2/3}$. For Keplerian rotation, ${\Omega }^{-2/3}=(GM{)}^{-1/3}r$. The positive numerical constant depends only on the dimensionless vertical profile and the convention for $\alpha$.

// Target: astrophysics.bigb

For a constant-opacity, gas-pressure-supported radiative disk surface at $z=H$, imposing $p=T=0$ does not impose $F=0$. If the outgoing surface flux is finite and positive, dividing hydrostatic balance by the temperature gradient gives $dp/dT\sim 16\sigma {\Omega }^{2}H{T}^3/(3\kappa F_s)$. Consequently $p\propto T^4$, $\rho \propto T^3$, and the diffusion equation gives $T\propto H-z$. Thus the mass density also vanishes regularly while the surface radiates a finite flux. This ideal boundary replaces the finite-temperature photospheric transition; its accuracy is not controlled solely by the disk aspect ratio.

// Target: astrophysics.bigb

In units $8\pi G=1$, a free homogeneous scalar has $\rho =P={\dot{\varphi }}_0^2/2$ and therefore a stiff equation of state. Its equation gives $a^2{\varphi }_0^{\prime }=\text{constant}$, while the Friedmann equation gives $3{\mathcal{H}}^2={\varphi }_0^{\prime 2}/2$. The displayed contracting branch satisfies both equations for negative conformal time. Constant normalizations and an additive scalar shift can be restored.

The conservative equation $u^{\prime \prime }=u^2-a^2$ has a saddle loop with $v_{+}(u)=\sqrt{2/3}(a-u)\sqrt{u+2a}$ for $-2a\le u\le a$. Its homoclinic orbit integrals are $I_0=2{\int }_{-2a}^a{v}_{+}(u)du$ and $I_1=2{\int }_{-2a}^{a}u{v}_{+}(u)du$. Substitution $s=u+2a$ evaluates them as displayed. For a weak perturbation $\epsilon (\beta +u)u^{\prime }$, integrating the energy derivative gives $\beta I_0+I_1=0$, hence $\beta =5a/7$. The nonzero derivative $I_0$ makes this a simple first-order splitting zero.

If the spectrum of an element $x$ lies in the open right half-plane, holomorphic functional calculus gives the displayed square root with spectrum in that half-plane. It is unique there. Two such roots $y,v$ have character values equal to the unique scalar square root with positive real part. Every character of an algebra is therefore nonzero on $y+v$, so the character description of the spectrum makes $y+v$ invertible. From $(y-v)(y+v)=0$ one obtains $y=v$. Equal character values alone do not prove equality in a possibly nonsemisimple algebra.

If the spectrum of an element $x$ avoids $(-\infty ,0]$, its holomorphic functional calculus value for the principal cube root is the unique $y$ with $y^3=x$ and spectrum in $\{z\ne 0:|\arg z|<\pi /3\}$. For uniqueness, any other such $v$ defines a continuous unital algebra homomorphism $f\mapsto {\Theta }_v(f\circ (z\mapsto z^3))$ from the slit-plane holomorphic functions to the algebra. It sends the coordinate to $x$, so continuity and uniqueness of holomorphic functional calculus identify it with ${\Theta }_x$. Applying it to the principal cube root gives $v=y$. This argument does not assume commutativity of the ambient algebra or separation by algebra characters.

The exponential series converges absolutely in a unital Banach algebra. Multiplication of absolutely convergent series gives $e^x{e}^{-x}=1$, and gives $e^{x+y}=e^x{e}^y$ when $x,y$ commute. Termwise differentiation gives $e^{tx}=1+tx+O(t^2)$ near zero. In a C-star algebra, if $h=h^{\ast }$ and $t$ is real, taking adjoints termwise gives $(e^{ith}{)}^{\ast }=e^{-ith}$, so this exponential is a Unitary element of a C-star algebra and has norm one. These facts let a norm bound on a functional recover its real values on Hermitian elements.

If a spectrum of an element splits into disjoint nonempty compact pieces $X,Y$, choose disjoint open neighborhoods and a holomorphic function equal to one on the first and zero on the second. The holomorphic functional calculus sends it to an idempotent with the corresponding zero/one values under every algebra character. It is neither zero nor identity: if one piece were killed, a reciprocal function on the other piece would make an originally spectral coordinate difference invertible.

If a compact spectrum of an element avoids zero and zero belongs to its unbounded complementary component, choose a simple arc from zero to infinity avoiding the Banach algebra spectrum. Its complement is a simply connected neighborhood of the Banach algebra spectrum, admitting a holomorphic logarithm. Apply holomorphic functional calculus to that logarithm and the composition rule for holomorphic functional calculus to the exponential to obtain the displayed identity. An invertible matrix over the complex numbers always satisfies the hypothesis because its finite Banach algebra spectrum cannot separate zero from infinity.

The Hölder exponent in Hölder continuity measures the power in the bound $|f(x)-f(y)|\le C|x-y{|}^{\alpha }$. For $0<\alpha \le 1$, a larger exponent imposes stronger local regularity on bounded distance ranges. Exponent one is Lipschitz continuity.

For a fixed positive sample size $n$ of independent and identically distributed Bernoulli random variables, letThe Hoeffding inequality makes $C_n$ a conservative confidence interval for $p$: ${\mathbb{P}}_p(p\in C_n)\ge 1-\delta$ for every $p\in [0,1]$. Indeed, exclusion from the closed interval implies $|{\hat{p}}_n-p|>r_n$, an event contained in the deviation event bounded by $\delta$. Clipping to $[0,1]$ retains coverage because $p$ already lies there. For $n=738$, $\delta =0.05$, and an observed frequency $369/738=1/2$, $r_n\approx 0.0499924$, giving approximately $[0.450008,0.549992]$.

Coverage concerns repeated independent samples: the unknown $p$ is fixed and the confidence interval is random. This fixed-$n$ guarantee does not justify repeatedly checking the same bound and stopping as soon as it looks favorable. The Hoeffding sample size for Bernoulli frequencies chooses $n$ in advance; guarantees across repeated inspection times require separate control of the additional events, for example through a union bound with an allocated failure probability at each time.

Let $B_1,\dots ,B_n$ be independent and identically distributed Bernoulli random variables with success probability $p$, and let ${\hat{p}}_n=n^{-1}{\sum }_i{B}_i$ be their observed frequency, a sample mean. Since every interval width is one, $V=n$. Substituting $t=n\epsilon$ into the Hoeffding inequality givesFor tolerance $0<\epsilon <1$ and failure probability $0<\delta <1$, solving $2e^{-2n{\epsilon }^2}\le \delta$ yields the sufficient sample sizeHere $\log$ is the natural logarithm. For $\epsilon =0.05$ and $\delta =0.05$, $n=738$ guarantees an error smaller than $0.05$ with probability at least $0.95$, for every $p\in [0,1]$. The two-sided bound at this sample size is approximately $0.0499440$. A specified single direction instead needs $n\ge \lceil \log (1/\delta )/(2{\epsilon }^2)\rceil$, or $600$ for these choices. Halving the tolerance multiplies the unrounded sufficient sample size by four; reducing the failure probability tenfold adds $\log (10)/(2{\epsilon }^2)$ before rounding.

This is a conservative sufficient sample size, not a necessary minimum. For a specified $p$, the count $K={\sum }_i{B}_i$ has a binomial distribution, whose exact upper tail probability isFor $k>np$, the Hoeffding inequality bounds this by $\exp [-2(k-np{)}^2/n]$. An arbitrary frequency threshold uses $k=\lceil n(p+\epsilon )\rceil$; if $k>n$, the event is impossible. The discreteness of the count and its variance $np(1-p)$ can make the exact tail probability substantially smaller. At $n=100$, $p=1/2$, $k=60$, the exact upper tail probability is about $0.0284440$, while the bound is $e^{-2}\approx 0.135335$. At $p=1/10$, $k=20$, the same deviation $\epsilon =0.1$ gives the same bound but an exact upper tail probability about $0.00197856$.

Let a surjective morphism from a smooth projective surface to a projective surface contract an irreducible curve $C$. Pull back an ample divisor $H$ from the target to obtain $L$. The projection formula for algebraic cycles gives $L\cdot C=0$ and $L^2=\mathrm{deg}(f)H^2>0$. The Hodge index theorem for algebraic surfaces makes the intersection form negative definite on the orthogonal complement of this positive-square class. Since an ample divisor on the source meets $C$ positively, its numerical class is nonzero, and $C^2<0$.